Completely forgotten FP2 integration...

Zygroth · 28-05-2009: 28th May 2009 22:39

Just started revising this chapter and I've realised I've pretty much forgotten all of it... I'll update this thread over the next few days with problems I can't solve and hopefully people will answer them, and maybe give tips on how to approach difficult integrations (these will come later!)

My working are in spoilers just to save space, because there are probably >100 Qs in the textbook and I doubt I can do many of them

$\displaystyle \int \mathrm{sech} 2x = \arctan e^{2x}$

Spoiler:

DeanK22 · 28-05-2009: 28th May 2009 23:33

sech(2x) = 1/cosh(2x) = ??

Spoiler:

Zygroth · 29-05-2009: 29th May 2009 00:26

Originally Posted by DeanK22

sech(2x) = 1/cosh(2x) = ??

Spoiler:

Multiply top and bottom by cosh 2x is how I did it...

Aah, that's how you do it, I didn't think to multiply sech 2x by e^2x so I was left with something I thought I couldn't integrate :/

Got stuck on $\displaystyle\int\dfrac{1}{2x^2 + 7x +3}$ - I can get a very similar answer to the $\dfrac{1}{5}\ln\dfrac{2x+1}{x+3}$ given

Spoiler:

Glutamic Acid · 29-05-2009: 29th May 2009 00:32

^ $\dfrac{1}{5} \ln \dfrac{x + 1/2}{x + 3} = \dfrac{1}{5} \ln \dfrac{1}{2} \dfrac{2x + 1}{x + 3} = \dfrac{1}{5} \ln \dfrac{2x+1}{x+3} + \dfrac{1}{5} \ln \dfrac{1}{2}$ , and the 1/5 ln(1/2) is sucked up into the arbitrary constant. Your integration is correct.

Zygroth · 29-05-2009: 29th May 2009 18:12

Thanks for the help so far

$\displaystyle\int x\mathrm{arcosh}x\ dx = 0.25(2x^2 - 1)\mathrm{arcosh}x - 0.25x\sqrt{x^2-1} + C$

Spoiler:

Glutamic Acid · 29-05-2009: 29th May 2009 19:47

x = cosh u would work.

Zygroth · 29-05-2009: 29th May 2009 20:52

Originally Posted by Glutamic Acid

x = cosh u would work.

To get $\int \sinh^2 u\ du$ ? After changing it to cosh 2u, integrating and expanding I get something quite complicated

That's using the arcosh x = ln ~ from the formula booklet.

Glutamic Acid · 29-05-2009: 29th May 2009 20:57

Originally Posted by Zygroth

To get $\int \sinh^2 u\ du$ ? After changing it to cosh 2u, integrating and expanding I get something quite complicated

That's using the arcosh x = ln ~ from the formula booklet.

That should work, but don't bother with any logarithms -- leave it in terms of inverse hyperbolic cosines.

Zygroth · 29-05-2009: 29th May 2009 23:00

Originally Posted by Glutamic Acid

That should work, but don't bother with any logarithms -- leave it in terms of inverse hyperbolic cosines.

But I had to show that the answer = the one on the right

Anyway, just doing:

$\displaystyle\int\frac{3-2x}{(3-x)^2} = -2\ln |x-3| + 3(x-3)^{-1}$

Spoiler:

Glutamic Acid · 29-05-2009: 29th May 2009 23:15

Originally Posted by Zygroth

But I had to show that the answer = the one on the right

Anyway, just doing:

$\displaystyle\int\frac{3-2x}{(3-x)^2} = -2\ln |x-3| + 3(x-3)^{-1}$

Spoiler:

$\dfrac{3 - 2x}{3 - x} = \dfrac{6 - 2x - 3}{3 - x} = 2 - \dfrac{3}{3 - x}$ , and the 2 is sucked in the arbitrary constant again.

Zygroth · 29-05-2009: 29th May 2009 23:54

Surprised it's taken this long to find an integration I have absolutely no clue on where to start, and it looks so, so easy...

$\dfrac{x+4}{x-4}$

Haven't looked at the answer yet, any tips?

Glutamic Acid · 29-05-2009: 29th May 2009 23:57

Rewrite it as $\dfrac{x - 4 + 8}{x - 4}$ , can you see what to do?

Gone Revising · 30-05-2009: 30th May 2009 00:24

Originally Posted by Zygroth

But I had to show that the answer = the one on the right

Anyway, just doing:

$\displaystyle\int\frac{3-2x}{(3-x)^2} = -2\ln |x-3| + 3(x-3)^{-1}$

Spoiler:

To get the answer on the right in a conventional way, you could use
partial fractions.

Zygroth · 30-05-2009: 30th May 2009 00:49

Originally Posted by Glutamic Acid

Rewrite it as $\dfrac{x - 4 + 8}{x - 4}$ , can you see what to do?

It was too easy to spot

I was expecting something more complicated

Seems I've forgotten how to integrate cos^4 x :/

Glutamic Acid · 30-05-2009: 30th May 2009 00:51

Originally Posted by Zygroth

It was too easy to spot

I was expecting something more complicated

Seems I've forgotten how to integrate cos^4 x :/

Try $\cos^4 x = (\cos^2 x)^2 = (\dfrac{1}{2}(1 + \cos 2x))^2$ ; repeating the double angle formula shebang later.

Zygroth · 30-05-2009: 30th May 2009 01:21

Originally Posted by Glutamic Acid

Try $\cos^4 x = (\cos^2 x)^2 = (\dfrac{1}{2}(1 + \cos 2x))^2$ ; repeating the double angle formula shebang later.

Hmm, I thought there might be some general way to integrate stuff like cos^n x

Just integrated cosech x/3, but it seems that ln tanh x/2 differentiates to cosech x - does anyone know of anything else like this I should learn? Cause there's no way I'd've been able to spot that in the exam

(Also, why isn't ln tan x/2 the integral of cosec x? :/ I tried it and it seemed fine...)

Glutamic Acid · 30-05-2009: 30th May 2009 01:27

Originally Posted by Zygroth

Hmm, I thought there might be some general way to integrate stuff like cos^n x

Just integrated cosech x/3, but it seems that ln tanh x/2 differentiates to cosech x - does anyone know of anything else like this I should learn? Cause there's no way I'd've been able to spot that in the exam

(Also, why isn't ln tan x/2 the integral of cosec x? :/ I tried it and it seemed fine...)

You can reduction formulae for cos^n; and usually even values of n are easier to integrate than odd ones. I wouldn't bother learning any actual integrals.

ln[tan(x/2)] is the integral of cosec x; if it says, in the formula booklet: - ln |cosec x + cot x| = $- \ln \left| \dfrac{1}{\sin x} + \dfrac{\cos x}{\sin x} \right| = - \ln \left| \dfrac{1 + \cos x}{\sin x} \right| = - \ln \left| \dfrac{2 \cos^2 (x/2)}{2 \cos (x/2) \sin (x/2)} \right| = - \ln |\cot(x/2)| = \ln |\tan(x/2)|$

Zygroth · 30-05-2009: 30th May 2009 01:39

Originally Posted by Glutamic Acid

~

Ah, I was wondering...Anyway I've finished with these "easy" questions and am now faced with ~15 pretty difficult (IMO) ones

My guess is I'll need help on all of them :/ Anyway:

Use the substitution u = 1/x to evaluate $\displaystyle \int^2_{\frac{2}{\sqrt 3}} \dfrac{1}{x\sqrt{x^2-1}}$

Trying to split it into partial fractions:

$\dfrac{A}{x} + \dfrac{B}{\sqrt{x^2-1}} \Rightarrow A\sqrt{x^2-1} + Bx = 1$
But I got no idea how to do it. If I let x = 1, then B = 1, but I don't know how to find A :/

Glutamic Acid · 30-05-2009: 30th May 2009 01:43

Originally Posted by Zygroth

Ah, I was wondering...Anyway I've finished with these "easy" questions and am now faced with ~15 pretty difficult (IMO) ones

My guess is I'll need help on all of them :/ Anyway:

Use the substitution u = 1/x to evaluate $\displaystyle \int^2_{\frac{2}{\sqrt 3}} \dfrac{1}{x\sqrt{x^2-1}}$

Trying to split it into partial fractions:

$\dfrac{A}{x} + \dfrac{B}{\sqrt{x^2-1}} \Rightarrow A\sqrt{x^2-1} + Bx = 1$
But I got no idea how to do it. If I let x = 1, then B = 1, but I don't know how to find A :/

One can't use partial fractions, it's not a rational function. You'll need to use the substitution given.

Zygroth · 30-05-2009: 30th May 2009 19:07

I need to use tan x/2 = t to integrate:

$\dfrac{1}{3+5\cos x}$ and $\dfrac{1}{5-3\cos x}$

I seem to remember having a lesson on using tan x/2 to integrate trig things, but I can't remember it :/

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Completely forgotten FP2 integration...